A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential en

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A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 72.0 J. What is the spring constant in this case?

Jun 16th, 2015

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Energy stored in spring is given by  E = (kx²)/2
therefore spring constant , k = 2.E /x²

where:
E is the energy stored  by spring = 72 J
x is the Displacement from equilibrium = 12 cm or  0.12 m

substituting values we get    k = 72* 2 /(0.12^2)  = 10000
Hence Spring constant  10000.0 kg s⁻²    (Answer)

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Jun 16th, 2015

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